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Can some one pls help with this integration watch

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    ∫( (e^x)/((e^x)-(e^-x))dx






    Thanks in advance!
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    e^x/( e^x - e^(-x) ) = e^2x / ( e^2x - 1 )

    So, the required integral is

    INT e^2x / ( e^2x - 1 ) dx

    You may be able to recognise that the numerator is half the derivitive of the denominator and then get the result straight away. If not, let u=e^2x, then x=(1/2)ln(u) and dx/du = 1/2u. Substitute these in

    INT e^2x / ( e^2x - 1 ) dx = INT e^2x / ( e^2x - 1 ) (dx/du) du = INT u/(u-1) (1/2u) du = (1/2) INT 1/(u-1) du = (1/2)ln|u-1| + C = (1/2)ln|e^2x - 1| + C

    I hope this helps.
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    ∫( (e^x)/((e^x)-(e^-x))dx
    couldnt you rewrite this as the integral of e^x/cosh x then use parts?

    what module is it from?
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    to me it looks like P3 (on edexcel at least), in which case hyperbolics aren't covered till P5
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    (Original post by rockindemon)
    ∫( (e^x)/((e^x)-(e^-x))dx
    couldnt you rewrite this as the integral of e^x/cosh x then use parts?

    what module is it from?
    e^x/(e^x-e^(-x)) = e^x/(2sinh(x))

    But also, this is P3 and I think my method above was simpler.
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    1/2 ln|e^2x-1| + C = 1/2 (C + x + ln|e^x - e^-x|)

    sorry, i just think that looks more aesthetic, especially as it leads on to a log of a hyperbolic sine, 1/2 (C + x + ln|sinh(x)|)
 
 
 

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